Tutor profile: Anubhav N.

If 20.0 mL of 0.100M HCl is titrated with 19.5 mL of NaOH solution, what is the molarity of the NaOH solution?

Firstly, it is important to note that at the end of this titration, an equivalence point is reached where the number of moles of acid will equal the number of moles of base. Let's find the number of moles of HCl. We know that molarity=(number of moles of solute/volume of solution), so number of moles of solute=(molarity*volume of solution). Thus, the number of moles of HCl will be (0.100M*0.020L)=0.002 moles. Note that the volume of solution has to be in L in standard units. This means that 0.002 moles of NaOH were added in the titration. We already know that 19.5 mL of NaOH was added. We can find the concentration of NaOH with the formula molarity=(number of moles of solute/volume of solution). Molarity of NaOH solution=(0.002 moles/0.0195L)=0.103M. Note again that we converted the volume of the NaOH solution to L to maintain standard units. The final answer is 0.103M.

Biologically, describe why we hiccup. What are some common triggers of hiccups?

A hiccup occurs due to a strong contraction of the diaphragm that is involuntary (not in our control). Soon after it contracts, we begin to inhale, but the epiglottis (cartilage attached to the opening of the larynx or windpipe) closes. Thus, the air from inhalation becomes trapped as it cannot move down the larynx and causes the "hic" sound. Hiccups are commonly triggered by acid reflux or when the phrenic nerve (nerve that innervates the diaphragm) is irritated.

Differentiate $$ y^{3}\ln(x) $$

To differentiate the product of two functions i.e. $$ y^{3} $$ and $$ \ln(x) $$ in this case, we have to use the product rule. According to this rule, the answer will be: [(first function*derivative of second function) + (second function*derivative of first function)]. Let's first find the derivative of $$ y^{3} $$. We know that the derivative of a general function $$ y^{n} $$ is $$ ny^{n-1} $$. Therefore, the derivative of $$ y^{3} $$ is $$ 3y^{2} $$. Next, let's find the derivative of $$ \ln(x) $$, which based on standard formulas for derivatives is $$ 1/x $$. Let's now substitute in the formula for the answer above: [(first function*derivative of second function) + (second function*derivative of first function)] = [($$ y^{3}*1/x $$) + ($$ \ln(x)*3y^{2} $$)] =$$ y^{3}/x $$ + ($$ 3y^{2}*\ln(x) $$)